What are the solutions to the equation t^2+4t-17=t ?

The solutions to the equation t^2+4t-17=t are t=(-3+sqrt(77))/2 ,t=(-3-sqrt(77))/2

Find the zeros of t^2+4t-17=t

The zeros of t^2+4t-17=t are t=(-3+sqrt(77))/2 ,t=(-3-sqrt(77))/2

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t^{2} + 4 t - 17 = t

t = \frac{- 3 + 77}{2}, t = \frac{- 3 - 77}{2}

Decimal Notation

t = 2.88748 \dots, t = - 5.88748 \dots

Solution steps

t^{2} + 4 t - 17 = t

Move

t

to the left side

t^{2} + 3 t - 17 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 17 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 17) = 77

t_{1, 2} = \frac{- 3 \pm 77}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 77}{2 \cdot 1}, t_{2} = \frac{- 3 - 77}{2 \cdot 1}

t = \frac{- 3 + 77}{2 \cdot 1} : \frac{- 3 + 77}{2}

t = \frac{- 3 - 77}{2 \cdot 1} : \frac{- 3 - 77}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 77}{2}, t = \frac{- 3 - 77}{2}

3 (x + 1)^{2} = 18

15 c^{2} - 8 c - 12 = 0

- 3 x^{2} + 15 x - 18 = 0

21 x^{2} - 13 - 20 = 0

\frac{1}{2} x^{2} + 3 x = 8

What are the solutions to the equation t^2+4t-17=t ?
The solutions to the equation t^2+4t-17=t are t=(-3+sqrt(77))/2 ,t=(-3-sqrt(77))/2
Find the zeros of t^2+4t-17=t
The zeros of t^2+4t-17=t are t=(-3+sqrt(77))/2 ,t=(-3-sqrt(77))/2